find the vertices of an equilaterall triangle circumscribed about the ellipse 9x^2 + 16y^2 = 144.

There can be two such triangles, One with apex up and one with apex down.

The ellipse has semi-major axis length of 4 and semi-minor axis length of 3.

\frac{x^{2}}{9}+\frac{y^{2}}{16}=1

Upper half of ellipse has equation:

y=\frac{3\sqrt{16-x^{2}}}{4}

y'=\frac{-3x}{4\sqrt{16-x^{2}}}

We know this slope must be -\sqrt{3}

because the equilateral triangle has angles of 60 degrees, and

tan(60)=\sqrt{3}. But the line on the right has negative slope.

\frac{-3x}{4\sqrt{16-x^{2}}}=-\sqrt{3}

x=\frac{16}{\sqrt{19}}.

Thus, the line is tangent to the ellipse at (\frac{16}{\sqrt{19}}, \;\ \frac{3\sqrt{57}}{19})

Using y-y_{1}=m(x-x_{1}) we can solve for the lower right vertex of the triangle.

\frac{3\sqrt{16-x^{2}}}{4}+3=-\sqrt{3}(x-a), where a is the x coordinate of this vertex.

Sub in the newly found x coordinate of the tangent point, x=\frac{16}{\sqrt{19}}, and we get:

\fbox{(\sqrt{3}+\sqrt{19}, \;\ -3)}

This means the left vertex has coordinates:

\fbox{(-\sqrt{3}-\sqrt{19}, \;\ -3)}

The apex has coordinate:

-3=-\sqrt{3}(\sqrt{3}+\sqrt{19})+b

b=\sqrt{57}

\fbox{(0, \;\ \sqrt{57})}

There is also a triangle that can be positioned with apex down as well.